How to solve partial differential equation Hi I have gotten a hard time solving this problem.
$$
\begin{array}{c}
    f(x,y) \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} = 1
    \\\\
    f(s,s) = \frac{s}{2} \quad , 0<s<1
\end{array}
$$
Looking for steps for this since I am trying to learn
 A: $$f(x,y) \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} = 1$$
The system of characteristic ODEs is :
$$\begin{cases}
\frac{dx}{dt}=f\\
\frac{dy}{dt}=1\\
\frac{df}{dt}=1\\
\end{cases}\quad\iff\quad \frac{dx}{f}=\frac{dy}{1}=\frac{df}{1}=dt$$
A first characteristic equation comes from $\frac{dx}{f}=\frac{df}{1}$
$$f^2-2x=c_1$$
A second characteristic equation comes from $\frac{dy}{1}=\frac{df}{1}$
$$f-y=c_2$$
General solution of the PDE on the form of implicit equation $c_1=\Phi(c_2)$ :
$$f^2-2x=\Phi(f-y)$$
$\Phi$ is an arbitrary function to be determined according to the specified condiion $f(s,s)=\frac{s}{2}$
$\left(\frac{s}{2}\right)^2-2s=\Phi(\frac{s}{2}-s)$ 
Let $X=-\frac{s}{2}\quad\implies\quad s=-2X$
$X^2+4X=\Phi(X)$
Now the function $\Phi$ is determined : 
$$\Phi(X)=X^2+4X$$
We put it into the above general solution where $X=f-y$ :
$$f^2-2x=(f-y)^2+4(f-y)$$
$$\boxed{f(x,y)=\frac{y-2}{2}+\frac{x-2}{y-2}}$$
This solution isn't valid for any $x,y$ since the condition $f(s,s)=\frac{s}{2}$ is limited to $0<s<1$
$-\frac12<X<0$
$-\frac12<f-y<0$
$-\frac12<\frac{-y-2}{2}+\frac{x-2}{y-2}<0$
$$\frac12<-\frac{y}{2}+\frac{x-2}{y-2}<1$$
From this one can find the range of $(x,y)$ where the above solution is valid with respect to the specified condition.
